Certify or not? An analysis of organic food supply chain with competing suppliers

Abstract

Customers expect companies to provide clear health-related information for the products they purchase in a big data environment. Organic food is data-enabled with the organic label, but the certification cost discourages small-scale suppliers from certifying their product. This lack of a label means that product that satisfies the organic standard is regarded as conventional product. By considering the trade-off between the profit gained from organic label and additional costs of certification, this paper investigates an organic food supply chain where a leading retailer procures from two suppliers with different brands. Customers care about both the brand-value and quality (more specifically, if food is organic or not) when purchasing the product. We explore the organic certification and wholesale pricing strategies for suppliers, and the supplier selection and retail pricing strategies for the retailer. We find that when two suppliers adopt asymmetric certification strategy, the retailer tends to procure the product with organic label. The supplier without a brand name can compensate with organic certification, which leads to more profits than the branded rival. As the risk of being abandoned by the retailer increases, the supplier without a brand name is more eager than the rival to obtain the organic label. If both suppliers certify the product, however, they will fall into a prisoner’s dilemma under situation with low health utility from organic label and high certification cost.

Appendices

Appendix 1: The equilibrium prices for a given certification strategy

We consider the situation of the retailer procuring two products, i.e., $\frac{{p_{2} - \upzeta_{2} \Delta }}{{p_{1} - \upzeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\upzeta_{1} - \upzeta_{2} } \right)\Delta .$

$\pi_{s1} = w_{1} \left( {1 - \frac{{w_{1} + m_{1} - \left( {w_{2} + m_{2} } \right) + \left( {\upzeta_{2} - \upzeta_{1} } \right)\Delta }}{1 - \theta }} \right) - \upzeta_{1} F$, $\pi_{s2} = w_{2} \left( {\frac{{w_{1} + m_{1} - \left( {w_{2} + m_{2} } \right) + \left( {\upzeta_{2} - \upzeta_{1} } \right)\Delta }}{1 - \theta } - \frac{{w_{2} + m_{2} - \upzeta_{2} \Delta }}{\theta }} \right) - \upzeta_{2} F$, let $\frac{{\partial \pi_{s1} }}{{\partial w_{1} }} = \frac{{\partial \pi_{s2} }}{{\partial w_{2} }} = 0$, we obtain $w_{1} = \frac{{2 - 2\theta - \left( {2 - \theta } \right)m_{1} + m_{2} + \left( {2 - \theta } \right)\upzeta_{1} \Delta - \upzeta_{2} \Delta }}{4 - \theta }$ and $w_{2} = \frac{{\theta m_{1} - \left( {2 - \theta } \right)m_{2} + \theta \left( {1 - \theta - \upzeta_{1} \Delta } \right) + \left( {2 - \theta } \right)\upzeta_{2} \Delta }}{4 - \theta }$. Substituting $w_{1}$ and $w_{2}$ into (5), we get: $\pi_{r} = m_{1} \left( {\frac{{2 - 2\theta - \left( {2 - \theta } \right)m_{1} + m_{2} + \left( {2 - \theta } \right)\upzeta_{1} \Delta - \upzeta_{2} \Delta }}{{\left( {4 - \theta } \right)\left( {1 - \theta } \right)}}} \right) + m_{2} \left( {\frac{{\theta m_{1} - \left( {2 - \theta } \right)m_{2} + \theta \left( {1 - \theta - \upzeta_{1} \Delta } \right) + \left( {2 - \theta } \right)\upzeta_{2} \Delta }}{{\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta }}} \right)$, let $\frac{{\partial \pi_{r} }}{{\partial m_{1} }} = \frac{{\partial \pi_{r} }}{{\partial m_{2} }} = 0$, we obtain $m_{2}^{*} = \frac{1}{2}\left( {\theta + \upzeta_{2} \Delta } \right)$, $m_{1}^{*} = \frac{1}{2}\left( {1 + \upzeta_{1} \Delta } \right)$. Replacing $m_{2}$ and $m_{1}$ with $m_{2}^{*}$ and $m_{1}^{*}$ into $w_{2}^{*}$ and $w_{1}^{*}$, then we have $w_{1}^{*} = \frac{{2 - 2\theta + \left( {2 - \theta } \right)\upzeta_{1} \Delta - \upzeta_{2} \Delta }}{{2\left( {4 - \theta } \right)}}$ and $w_{2}^{*} = \frac{{\theta \left( {1 - \theta - \upzeta_{1} \Delta } \right) + \left( {2 - \theta } \right)\upzeta_{2} \Delta }}{{2\left( {4 - \theta } \right)}}$. We further get Eqs. (15)–(19).

Appendix 2: Certification strategy if $\theta = 1$

If $\theta = 1$, the utility a customer derives from product $n$ can be denoted as $u_{n} = v - p_{n} + \upzeta_{n} \Delta$, the equilibrium price will satisfy $p_{2} - p_{1} = \left( {\upzeta_{2} - \upzeta_{1} } \right)\Delta$. If not, the demand of one supplier will be zero, so this supplier will adjust his wholesale price to satisfy $p_{2} - p_{1} = \left( {\upzeta_{2} - \upzeta_{1} } \right)\Delta$. The customer will purchase the product if $u_{n} \ge 0$, the overall demand will be $D = 1 - p_{n} + \upzeta_{n} \Delta$, and the final demand for supplier $n$ will be split equally, which can be denoted as $D_{n} = \frac{{1 - p_{n} + \upzeta_{n} \Delta }}{2}$. By maximizing $\pi_{sn} = \left( {\frac{{1 - \left( {w_{n} + m_{n} } \right) + \upzeta_{n} \Delta }}{2}} \right)w_{n}$, we get $w_{n}^{*} = \frac{1}{2}\left( {1 - m_{n} + \varDelta \upzeta_{n} } \right)$. Substituting $w_{n}^{*}$ into $\pi_{r} = \mathop \sum \nolimits_{n = 1}^{2} \left( {\frac{{1 - \left( {w_{n} + m_{n} } \right) + \upzeta_{n} \Delta }}{2}} \right)m_{n}$ and making $\frac{{\partial \pi_{r} }}{{\partial m_{n} }} = 0$, we get $m_{n}^{*} = \frac{1}{2}\left( {1 + \varDelta \upzeta_{n} } \right)$. Substituting $m_{n}^{*}$ into $w_{n}^{*}$, we get $w_{n}^{*} = \frac{1}{4}\left( {1 + \varDelta \upzeta_{n} } \right)$ and $p_{n}^{*} = \frac{3}{4}\left( {1 + \varDelta \upzeta_{n} } \right)$. By substituting $p_{n}^{*}$ into the constraint of $p_{2} - p_{1} = \left( {\upzeta_{2} - \upzeta_{1} } \right)\Delta$, we get $\upzeta_{1} = \upzeta_{2}$. The suppliers will certify the product if $\pi_{sn} \left( {\upzeta_{n} = 1} \right) - \pi_{sn} \left( {\upzeta_{n} = 0} \right) > 0$, i.e., $F < \frac{1}{32}\varDelta \left( {2 + \varDelta } \right)$.

Appendix 3

Proof of Lemma 1

• (i) $\frac{{\partial w_{1}^{*} }}{\partial \theta } = - \frac{{6 + 2\varDelta \zeta_{1} + \varDelta \zeta_{2} }}{{2\left( {4 - \theta } \right)^{2} }} < 0$,

  $$ \frac{{\partial \pi_{s1}^{*} }}{\partial \theta } = \frac{{\left( {2 - 2\theta + \varDelta \left( {2 - \theta } \right)\zeta_{1} - \varDelta \zeta_{2} } \right)\left( {2\left( { - 2 + \theta + \theta^{2} } \right) + \varDelta \left( {4 - \left( {2 - \theta } \right)\theta } \right)\zeta_{1} - 3\varDelta \left( {2 - \theta } \right)\zeta_{2} } \right)}}{{4\left( {4 - \theta } \right)^{3} \left( {1 - \theta } \right)^{2} }} < 0. $$

• (ii) $\frac{{\partial w_{2}^{*} }}{\partial \theta } = \frac{{4 - \left( {8 - \theta } \right)\theta - 2\varDelta \left( {2\zeta_{1} + \zeta_{2} } \right)}}{{2\left( {4 - \theta } \right)^{2} }}$, when $\varDelta = 0$, the maximum of $\frac{{\partial w_{2}^{*} }}{\partial \theta }$ equals $\frac{{4 - \left( {8 - \theta } \right)\theta }}{{2\left( {4 - \theta } \right)^{2} }}$, thus, we find that when $\theta > 4 - 2\sqrt 3$, $\frac{{\partial w_{2}^{*} }}{\partial \theta } < 0$, when $\theta < 4 - 2\sqrt 3$, if $\varDelta < \frac{{4 - 8\theta + \theta^{2} }}{{4\zeta_{1} + 2\zeta_{2} }}$, $\frac{{\partial w_{2}^{*} }}{\partial \theta } > 0$, else, $\frac{{\partial w_{2}^{*} }}{\partial \theta } < 0$.

• (iii) In scenario I, ${\upzeta }_{1} = {\upzeta }_{2} = 1$, after substituting $p_{1}^{*}$ and $p_{2}^{*}$ into the constraint $\frac{{p_{2} - \upzeta_{2} \Delta }}{{p_{1} - \upzeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\upzeta_{1} - \upzeta_{2} } \right)\Delta$, we get that the retailer will procure two products simultaneously. In scenario II, $\upzeta_{1} = 1$, $\upzeta_{2} = 0$, after substituting $p_{1}^{*}$ and $p_{2}^{*}$ into the constraint $\frac{{p_{2} - \upzeta_{2} \Delta }}{{p_{1} - \upzeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\upzeta_{1} - \upzeta_{2} } \right)\Delta$, we get that the retailer will procure two products simultaneously if $\varDelta \le 1 - \theta$; otherwise, he will only procure products from supplier 1.

  $$ \begin{aligned} \pi_{s2}^{*} & = \frac{{\left( {\theta \left( { - 1 + \theta + \varDelta } \right) - \left( {2 - \theta } \right)\zeta_{2} \varDelta } \right)^{2} }}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }} - \zeta_{2} F,\quad \frac{{\partial \pi_{s2}^{*} }}{\partial \theta } = \frac{{\left( {\theta \left( {1 - \varDelta - \theta } \right) + \varDelta \left( {2 - \theta } \right)\zeta_{2} } \right)f_{1} \left( \varDelta \right)}}{{4\left( {4 - \theta } \right)^{3} \left( {1 - \theta } \right)^{2} \theta^{2} }}, \\ f_{1} \left( \varDelta \right) & = 4\theta - 11\theta^{2} + 7\theta^{3} + \varDelta \left( {18\theta \zeta_{2} + 2\theta^{3} \zeta_{2} + 2\theta^{3} - 4\theta - \theta^{2} - 9\theta^{2} \zeta_{2} - 8\zeta_{2} } \right). \\ \end{aligned} $$

  Since $\frac{{\partial f_{1} \left( \varDelta \right)}}{\partial \varDelta } < 0$, the maximum of $f_{1} \left( \varDelta \right)$ which equals $4\theta - 11\theta^{2} + 7\theta^{3}$ is obtained at $\varDelta = 0$. The minimum of $f_{1} \left( \varDelta \right)$ which equals $\left( { - 1 + \theta } \right)\left( {8 + \theta \left( { - 10 + 11\theta } \right)} \right)$ in scenario I and $- 2\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta^{2}$ in scenario II is obtained at $\varDelta = 1$ and $\varDelta = 1 - \theta$ respectively. We find that the minimum of $f_{1} \left( \varDelta \right)$ in scenario I and II is less than 0. Thus, when $\theta > \frac{4}{7}$, $f_{1} \left( {\varDelta = 0} \right) < 0$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$, when $\theta < \frac{4}{7}$, if $\varDelta < \frac{{\left( {1 - \theta } \right)\theta \left( { - 4 + 7\theta } \right)}}{{\theta \left( { - 4 + \theta \left( { - 1 + 2\theta } \right)} \right) + \left( { - 8 + \theta \left( {18 + \theta \left( { - 9 + 2\theta } \right)} \right)} \right)\zeta_{2} }}$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } > 0$, else, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$.

  In scenario I, $p_{1}^{*} - p_{2}^{*} = \frac{{\left( {1 - \theta } \right)\left( {6 - \varDelta - 2\theta } \right)}}{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial p_{1}^{*} - p_{2}^{*} }}{\partial \varDelta } = \frac{1 - \theta }{{2\left( { - 4 + \theta } \right)}} < 0$, $\frac{{\partial p_{1}^{*} - p_{2}^{*} }}{\partial \theta } = \frac{{3\left( {2 + \varDelta } \right) - 2\left( {4 - \theta } \right)^{2} }}{{2\left( {4 - \theta } \right)^{2} }} < 0$. $D_{1}^{*} = \frac{2 + \varDelta }{8 - 2\theta }$, $\frac{{\partial D_{1}^{*} }}{\partial \theta } > 0$, $D_{2}^{*} = \frac{2\varDelta + \theta }{{8\theta - 2\theta^{2} }}$, if $\Delta < \frac{{{\uptheta}^{2} }}{{8 - 4{\uptheta}}}$, $\frac{{\partial D_{2}^{*} }}{\partial \theta } > 0$, else, $\frac{{\partial D_{2}^{*} }}{\partial \theta } < 0$. $D_{1}^{*} - D_{2}^{*} = \frac{{\varDelta \left( { - 2 + \theta } \right) + \theta }}{{2\left( {4 - \theta } \right)\theta }}$, if $\theta > \frac{2\varDelta }{1 + \varDelta }$, $D_{1}^{*} > D_{2}^{*}$, else, $D_{1}^{*} < D_{2}^{*}$, $\frac{{\partial D_{1}^{*} - D_{2}^{*} }}{\partial \theta } = \frac{1}{4}\left( {\frac{2 + \varDelta }{{\left( {4 - \theta } \right)^{2} }} + \frac{\varDelta }{{\theta^{2} }}} \right) > 0$.

• (iv) In scenario III, ${\upzeta }_{1} = 0$, ${\upzeta }_{2} = 1$, $\pi_{s2}^{*} = \frac{{\left( {\theta \left( {1 - \theta } \right) + \left( {2 - \theta } \right)\Delta } \right)^{2} }}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }} - \zeta_{2} F$. After substituting $p_{1}^{*}$ and $p_{2}^{*}$ into the constraint $\frac{{p_{2} - \zeta_{2} \Delta }}{{p_{1} - \zeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\zeta_{1} - \zeta_{2} } \right)\Delta$, we get that the retailer will procure two products simultaneously if $\varDelta \le 2 - 2\theta$; otherwise, he will only procure products from supplier 2.

  $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } = \frac{{\left( {\varDelta \left( {2 - \theta } \right) + \left( {1 - \theta } \right)\theta } \right)f_{2} \left( \varDelta \right)}}{{4\left( {4 - \theta } \right)^{3} \left( {1 - \theta } \right)^{2} \theta^{2} }}$, $f_{2} \left( \varDelta \right) = \left( {1 - \theta } \right)\theta \left( {4 - 7\theta } \right) + \varDelta \left( { - 8 + \theta \left( {18 + \theta \left( { - 9 + 2\theta } \right)} \right)} \right)$, $f_{2} \left( {\varDelta = 0} \right) = \left( {1 - \theta } \right)\theta \left( {4 - 7\theta } \right)$, if $\theta > \frac{4}{7}$, $f_{2} \left( {\varDelta = 0} \right) < 0$, else, $f_{2} \left( {\varDelta = 0} \right) > 0$. $f_{2} \left( {\varDelta = 2 - 2\theta } \right) = - \left( {4 - \theta } \right)\left( {1 - \theta } \right)\left( {4 - \theta \left( {9 - 4\theta } \right)} \right)$, if $\theta > \frac{{9 - \sqrt {17} }}{8}$, $f_{2} \left( {\varDelta = 2 - 2\theta } \right) > 0$, else, $f_{2} \left( {\varDelta = 2 - 2\theta } \right) < 0$. Thus, we get that when $\theta < \frac{4}{7}$, if $\varDelta < \frac{{\left( {1 - \theta } \right)\theta \left( { - 4 + 7\theta } \right)}}{{ - 8 + \theta \left( {18 + \theta \left( { - 9 + 2\theta } \right)} \right)}}$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } > 0$, else, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$; when $\frac{4}{7} < \theta < \frac{{9 - \sqrt {17} }}{8}$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$; when $\theta > \frac{{9 - \sqrt {17} }}{8}$, if $\varDelta < \frac{{\left( {1 - \theta } \right)\theta \left( { - 4 + 7\theta } \right)}}{{ - 8 + \theta \left( {18 + \theta \left( { - 9 + 2\theta } \right)} \right)}}$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$, else, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } > 0$.

  In scenario IV, ${\upzeta }_{1} = {\upzeta }_{2} = 0$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } = \frac{4 - 7\theta }{{4\left( {4 - \theta } \right)^{3} }}$, if $\theta < \frac{4}{7}$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } > 0$, else, $\frac{{\partial \pi_{s2}^{*} }}{\partial \theta } < 0$.□

Proof of Proposition 1

• (i) $\frac{{\partial \pi_{r}^{ *} }}{\partial \theta } = \frac{{f_{3} \left( \varDelta \right)}}{{8\left( {4 - \theta } \right)^{2} \theta^{2} }}$, $f_{3} \left( \varDelta \right) = \varDelta^{2} \left( {8\theta - 16 + 2\theta^{2} } \right) + 12\varDelta \theta^{2} + 12\theta^{2}$, $f_{3} \left( {\varDelta = 1} \right) = - 16 + 8\theta + 26\theta^{2}$, if $\theta > \frac{2}{13}\left( {3\sqrt 3 - 1} \right)$, $f_{3} \left( {\varDelta = 1} \right) > 0$, thus, $\frac{{\partial \pi_{r}^{ *} }}{\partial \theta } > 0$; if $\theta < \frac{2}{13}\left( {3\sqrt 3 - 1} \right)$, $f_{3} \left( {\varDelta = 1} \right) < 0$, if $\varDelta < \varDelta_{r}^{*}$, $f_{3} \left( \varDelta \right) > 0$, thus, $\frac{{\partial \pi_{r}^{ *} }}{\partial \theta } > 0$; if $\varDelta > \varDelta_{r}^{*}$, $\frac{{\partial \pi_{r}^{ *} }}{\partial \theta } < 0$.

• (ii) $\frac{{\partial \pi_{sc}^{ *} }}{\partial \theta } = \frac{{f_{4} \left( \varDelta \right)}}{{4\left( {4 - \theta } \right)^{3} \theta^{2} }}$, $f_{4} \left( \varDelta \right) = 2\varDelta \theta^{2} \left( {4 - 7\theta } \right) + \theta^{2} \left( {20 - 17\theta } \right) + 2\varDelta^{2} \left( {\theta \left( {18 - \theta \left( {5 + \theta } \right)} \right) - 24} \right)$, $f_{4} \left( {\varDelta = 1} \right) = - 48 + 3\theta \left( {12 + \left( {6 - 11\theta } \right)\theta } \right) < 0$, if $\varDelta < \frac{{\left( {4 - 7\theta } \right)\theta^{2} + \theta \sqrt {3\left( { - 4 + \theta } \right)^{2} \left( {20 - \theta \left( {22 - 5\theta } \right)} \right)} }}{{2\left( {24 + \theta \left( { - 18 + \theta \left( {5 + \theta } \right)} \right)} \right)}}$, $\frac{{\partial \pi_{sc}^{ *} }}{\partial \theta } > 0$, else, $\frac{{\partial \pi_{sc}^{ *} }}{\partial \theta } < 0$.

• (iii) $E\left( {CS} \right) = \frac{{2\varDelta \theta \left( {8 + \theta } \right) + \left( {\varDelta^{2} + \theta } \right)\left( {4 + 5\theta } \right)}}{{8\left( {4 - \theta } \right)^{2} \theta }}$, $\frac{{\partial E\left( {CS} \right)}}{\partial \theta } = \frac{{f_{5} \left( \varDelta \right)}}{{8\left( {4 - \theta } \right)^{3} \theta^{2} }}$, $f_{5} \left( \varDelta \right) = 2\varDelta \theta^{2} \left( {20 + \theta } \right) + 2\varDelta^{2} \left( {2 + \theta } \right)\left( { - 4 + 5\theta } \right) + \theta^{2} \left( {28 + 5\theta } \right)$, $f_{5} \left( {\varDelta = 1} \right) = - 16 + \theta \left( {12 + \theta \left( {78 + 7\theta } \right)} \right)$, when $\theta > \frac{4}{5}$, $f_{5} \left( \varDelta \right) > 0$, when $\theta < \frac{4}{5}$, we let $- 16 + \theta_{cs}^{*} \left( {12 + \theta_{cs}^{*} \left( {78 + 7\theta_{cs}^{*} } \right)} \right) = 0$, if $\theta < \theta_{cs}^{*}$, $f_{5} \left( {\varDelta = 1} \right) < 0$, if $\theta > \theta_{cs}^{*}$, $f_{5} \left( {\varDelta = 1} \right) > 0$. Thus, we get if $\theta > \theta_{cs}^{*}$, $\frac{{\partial E\left( {CS} \right)}}{\partial \theta } > 0$; otherwise, it increases in $\theta$ when $\varDelta < \varDelta_{cs}^{*}$ and decreases in $\theta$ when $\varDelta > \varDelta_{cs}^{*}$.

Proof of Proposition 2

Since $D_{1}^{*} - D_{2}^{*} = \frac{{\theta - \varDelta \left( {2 - \theta } \right)}}{{2\left( {4 - \theta } \right)\theta }}$, we get if $\varDelta < \frac{\theta }{2 - \theta }$, $D_{1}^{*} > D_{2}^{*}$, if $\varDelta > \frac{\theta }{2 - \theta }$, $D_{1}^{*} < D_{2}^{*}$, $\frac{{\partial D_{1}^{*} - D_{2}^{*} }}{\partial \varDelta } = \frac{{ - \left( {2 - \theta } \right)}}{{2\left( {4 - \theta } \right)\theta }} < 0$, $\frac{{\partial D_{1}^{*} - D_{2}^{*} }}{\partial \theta } = \frac{1}{4}\left( {\frac{2 + \varDelta }{{\left( {4 - \theta } \right)^{2} }} + \frac{\varDelta }{{\theta^{2} }}} \right) > 0$. $\frac{{\partial \pi_{s1}^{*} }}{{\partial\varDelta }} = \frac{{\left( {2 + \varDelta } \right)\left( {1 - \theta } \right)}}{{2\left( {4 - \theta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{s2}^{*} }}{{\partial\varDelta }} = \frac{{\left( {1 - \theta } \right)\left( {2\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }} > 0$, $\pi_{s1}^{*} - \pi_{s2}^{*} = \frac{{\left( {\varDelta^{2} - \theta } \right)\left( {1 - \theta } \right)}}{{4\left( { - 4 + \theta } \right)\theta }}$, if $\varDelta > \sqrt \theta$, $\pi_{s1}^{*} < \pi_{s2}^{*}$, if $\varDelta < \sqrt \theta$, $\pi_{s1}^{*} > \pi_{s2}^{*}$. $\frac{{\partial \pi_{s1}^{*} }}{{\partial\varDelta }} - \frac{{\partial \pi_{s2}^{*} }}{{\partial\varDelta }} = \frac{{\varDelta \left( {1 - \theta } \right)}}{{2\left( { - 4 + \theta } \right)\theta }} < 0$. $p_{1}^{*} - p_{2}^{*} = \frac{{\left( {1 - \theta } \right)\left( {6 - \varDelta - 2\theta } \right)}}{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial p_{1}^{*} - p_{2}^{*} }}{\partial \varDelta } = - \frac{{\left( {1 - \theta } \right)}}{{2\left( {4 - \theta } \right)}} < 0$.

Compared with the scenario in which neither certify the product, the increase of supplier 1’s profit can be denoted by $\pi_{{\Delta s1}}^{*} = \pi_{s1}^{*} \left( {\delta_{1} = \delta_{2} = 1} \right) - \pi_{s1}^{*} \left( {\delta_{1} = \delta_{2} = 0} \right) = \frac{{\varDelta \left( {4 + \varDelta } \right)\left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} }}$, the increase of supplier 2’s profit can be denoted by $\pi_{{\Delta s2}}^{*} = \pi_{s2}^{*} \left( {\delta_{1} = \delta_{2} = 1} \right) - \pi_{s2}^{*} \left( {\delta_{1} = \delta_{2} = 0} \right) = \frac{{\varDelta \left( {1 - \theta } \right)\left( {\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }}$. $\frac{{\partial \pi_{{\Delta s1}}^{*} }}{\partial \theta } = \frac{{\varDelta \left( {4 + \varDelta } \right)\left( {2 + \theta } \right)}}{{4\left( { - 4 + \theta } \right)^{3} }} < 0$, $\frac{{\partial \pi_{{\Delta s2}}^{*} }}{\partial \theta } = \frac{{\varDelta \left( {\theta^{2} \left( {2 + \theta } \right) + \varDelta \left( {4 - \theta \left( {3 - 2\theta } \right)} \right)} \right)}}{{\left( { - 4 + \theta } \right)^{3} \theta^{2} }} < 0$.

Proof of Proposition 3

When $\delta_{1} = \delta_{2} = 1$ or $\delta_{1} = \delta_{2} = 0$, we find that $p_{1}^{*}$ and $p_{2}^{*}$ always satisfy $\frac{{p_{2} - {\upzeta }_{2}\Delta }}{{p_{1} - {\upzeta }_{1}\Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {{\upzeta }_{1} - {\upzeta }_{2} } \right)\Delta$; when $\delta_{1} = 1$, $\delta_{2} = 0$, $p_{1}^{*}$ and $p_{2}^{*}$ will satisfy the above constraint if $\varDelta \le 1 - \theta$; when $\delta_{1} = 0$, $\delta_{2} = 1$, $p_{1}^{*}$ and $p_{2}^{*}$ will satisfy the above constraint if $\varDelta \le 2 - 2\theta$.

When $\delta_{1} = \delta_{2} = 1$, $\frac{{\partial w_{1}^{*} }}{\partial \varDelta } = \frac{1 - \theta }{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial w_{2}^{*} }}{\partial \varDelta } = \frac{1 - \theta }{{\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial p_{1}^{*} }}{\partial \varDelta } = \frac{5 - 2\theta }{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial p_{2}^{*} }}{\partial \varDelta } = \frac{6 - 3\theta }{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial \pi_{s1}^{*} }}{\partial \varDelta } = \frac{2 - 2\theta + \varDelta - \varDelta \theta }{{2\left( { - 4 + \theta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \varDelta } = \frac{{\varDelta \left( {2 - \theta } \right) + \theta \left( {1 - \varDelta - \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }} > 0$, $\frac{{\partial \pi_{r}^{*} }}{\partial \varDelta } = \frac{{\varDelta \left( {2 - \theta } \right)\theta + 2\theta \left( {1 - \theta - \varDelta } \right) + \left( {\left( {1 - \theta } \right)\theta + \varDelta \left( {2 - \theta } \right)} \right)}}{{2\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta }} > 0$; When $\delta_{1} = 1$, $\delta_{2} = 0$, $\frac{{\partial w_{1}^{*} }}{\partial \varDelta } = \frac{{\left( {2 - \theta } \right)}}{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial w_{2}^{*} }}{\partial \varDelta } = \frac{\theta }{{2\left( { - 4 + \theta } \right)}} < 0$, $\frac{{\partial p_{1}^{*} }}{\partial \varDelta } = \frac{3 - \theta }{4 - \theta } > 0$, $\frac{{\partial p_{2}^{*} }}{\partial \varDelta } = \frac{\theta }{{2\left( { - 4 + \theta } \right)}} < 0$, $\frac{{\partial \pi_{s1}^{*} }}{\partial \varDelta } = \frac{{\left( {2 - \theta } \right)\left( {2 - 2\theta - \varDelta \left( {2 - \theta } \right)} \right)}}{{2\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} > 0$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \varDelta } = \frac{{\left( {\theta \left( { - 1 + \theta + \varDelta } \right)} \right)}}{{2\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} < 0$, $\frac{{\partial \pi_{r}^{*} }}{\partial \varDelta } = \frac{{\varDelta \left( {2 - \theta } \right)\theta + 2\theta \left( {1 - \theta } \right)}}{{2\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta }} > 0$; When $\delta_{1} = 0$, $\delta_{2} = 1$, $\frac{{\partial w_{1}^{*} }}{\partial \varDelta } = \frac{1}{{2\left( { - 4 + \theta } \right)}} < 0$, $\frac{{\partial w_{2}^{*} }}{\partial \varDelta } = \frac{2 - \theta }{{2\left( {4 - \theta } \right)}} > 0$, $\frac{{\partial p_{1}^{*} }}{\partial \varDelta } = \frac{1}{{2\left( { - 4 + \theta } \right)}} < 0$, $\frac{{\partial p_{2}^{*} }}{\partial \varDelta } = \frac{3 - \theta }{4 - \theta } > 0$, $\frac{{\partial \pi_{s1}^{*} }}{\partial \varDelta } = \frac{{\left( { - 2 + 2\theta + \varDelta } \right)}}{{2\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} < 0$, $\frac{{\partial \pi_{s2}^{*} }}{\partial \varDelta } = \frac{{\left( {2 - \theta } \right)\left( {\theta \left( {1 - \theta } \right) + \varDelta \left( {2 - \theta } \right)} \right)}}{{2\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }} > 0$, $\frac{{\partial \pi_{r}^{*} }}{\partial \varDelta } = \frac{{\varDelta \left( {2 - \theta } \right)\theta + 2\theta \left( {1 - \theta } \right)}}{{2\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta }} > 0$.

Proof of Proposition 4

We first consider the condition that $\varDelta \le 1 - \theta$, the retailer will procure the products from both suppliers regardless of their organic label. To simplify the exposition, we use $\pi_{s2yy}^{*}$ to represent supplier 2’s profit in the certification–certification scenario; similar symbols can be used to represent the two suppliers’ profit in different certification scenarios.

For supplier 2, given that supplier 1 adopts certification, supplier 2 adopts certification if $\pi_{s2yy}^{*} > \pi_{s2yn}^{*}$, otherwise no certification; given that supplier 1 adopts no certification, supplier 2 adopts certification if $\pi_{s2ny}^{*} > \pi_{s2nn}^{*}$, otherwise no certification. For supplier 1, given that supplier 2 adopts certification, supplier 1 adopts certification if $\pi_{s1yy}^{*} > \pi_{s1ny}^{*}$, otherwise no certification; given that supplier 2 adopts no certification, supplier 1 adopts certification if $\pi_{s1yn}^{*} > \pi_{s1nn}^{*}$, otherwise no certification.

By equating $\pi_{s2yy}^{*}$ and $\pi_{s2yn}^{*}$, we get the threshold certification cost $\bar{F}_{11} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {2\left( {1 - \theta } \right)\theta + \varDelta \left( {2 - 3\theta } \right)} \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$; by equating $\pi_{s2ny}^{*}$ and $\pi_{s2nn}^{*}$, we get $\bar{F}_{12} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$; by equating $\pi_{s1yy}^{*}$ and $\pi_{s1ny}^{*}$, we get $\bar{F}_{13} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {4 - \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}}$; by equating $\pi_{s1yn}^{*}$ and $\pi_{s1nn}^{*}$, we get $\bar{F}_{14} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {2\left( {2 + \varDelta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}}$.

Since $\frac{{\bar{F}_{12} }}{{\bar{F}_{13} }} = \frac{{\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta }}{{\left( {4 - \left( {4 + \varDelta } \right)\theta } \right)\theta }}$, we get if $\varDelta < \varDelta_{11} = - 2 + \frac{4}{{2 + \left( { - 1 + \theta } \right)\theta }}$, $\bar{F}_{12} < \bar{F}_{13}$, else, $\bar{F}_{12} > \bar{F}_{13}$. Since $\frac{{\bar{F}_{11} }}{{\bar{F}_{13} }} = \frac{{\left( {2\left( {1 - \theta } \right)\theta + \varDelta \left( {2 - 3\theta } \right)} \right)}}{{\theta \left( {4 - \left( {4 + \varDelta } \right)\theta } \right)}}$, we get when $\theta > \frac{{5 - \sqrt {17} }}{2} \approx 0.438$, $\bar{F}_{11} < \bar{F}_{13}$, when $\theta < \frac{{5 - \sqrt {17} }}{2}$, if $\varDelta < \varDelta_{12} = \frac{2\theta }{2 - \theta }$, $\bar{F}_{11} < \bar{F}_{13}$, else if $\varDelta > \varDelta_{12}$, $\bar{F}_{11} > \bar{F}_{13}$. Since $\frac{{\bar{F}_{12} }}{{\bar{F}_{14} }} = \frac{{\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta }}{{\theta \left( {2\left( {2 + \varDelta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}$, we get when $\theta > \frac{{5 - \sqrt {17} }}{2}$, $\bar{F}_{12} < \bar{F}_{14}$, when $\theta < \frac{{5 - \sqrt {17} }}{2}$, if $\varDelta < \varDelta_{12} = \frac{2\theta }{2 - \theta }$, $\bar{F}_{12} < \bar{F}_{14}$, else if $\varDelta > \varDelta_{12}$, $\bar{F}_{12} > \bar{F}_{14}$. Since $\frac{{\bar{F}_{11} }}{{\bar{F}_{14} }} = \frac{{\left( {2\left( {1 - \theta } \right)\theta + \varDelta \left( {2 - 3\theta } \right)} \right)}}{{\theta \left( {2\left( {2 + \varDelta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}$, we get when $\theta > \frac{{7 - \sqrt {41} }}{2} \approx 0.298$, $\bar{F}_{11} < \bar{F}_{14}$, when $\theta < \frac{{7 - \sqrt {41} }}{2}$, if $\varDelta < \varDelta_{13} = \frac{{2\left( {1 - \theta } \right)\theta }}{{2 - \left( {5 - \theta } \right)\theta }}$, $\bar{F}_{11} < \bar{F}_{14}$, else if $\varDelta > \varDelta_{13}$, $\bar{F}_{11} > \bar{F}_{14}$. In conclusion, we obtain the following results:

In the region of $\theta < \frac{{7 - \sqrt {41} }}{2}$: when $\varDelta < \varDelta_{11}$, $\bar{F}_{11} < \bar{F}_{12} < \bar{F}_{13} < \bar{F}_{14}$; when $\varDelta_{11} < \varDelta < \varDelta_{12}$, $\bar{F}_{11} < \bar{F}_{13} < \bar{F}_{12} < \bar{F}_{14}$; when $\varDelta_{12} < \varDelta < \varDelta_{13}$, $\bar{F}_{13} < \bar{F}_{11} < \bar{F}_{14} < \bar{F}_{12}$; when $\varDelta > \varDelta_{13}$, $\bar{F}_{13} < \bar{F}_{14} < \bar{F}_{11} < \bar{F}_{12}$.

In the region of $\frac{{7 - \sqrt {41} }}{2} < \theta < \frac{{5 - \sqrt {17} }}{2}$: when $\varDelta < \varDelta_{11}$, $\bar{F}_{11} < \bar{F}_{12} < \bar{F}_{13} < \bar{F}_{14}$; when $\varDelta_{11} < \varDelta < \varDelta_{12}$, $\bar{F}_{11} < \bar{F}_{13} < \bar{F}_{12} < \bar{F}_{14}$; when $\varDelta > \varDelta_{12}$, $\bar{F}_{13} < \bar{F}_{11} < \bar{F}_{14} < \bar{F}_{12}$.

In the region of $\theta > \frac{{5 - \sqrt {17} }}{2}$: when $\varDelta < \varDelta_{11}$, $\bar{F}_{11} < \bar{F}_{12} < \bar{F}_{13} < \bar{F}_{14}$, when $\varDelta > \varDelta_{11}$, $\bar{F}_{11} < \bar{F}_{13} < \bar{F}_{12} < \bar{F}_{14}$.

In conclusion, we obtain the results in Table 2.

For supplier 1, the benefit from certification can be denoted as $\pi_{s1}^{*} \left( {{\upzeta }_{1} = {\upzeta }_{2} = 1} \right) - \pi_{s1}^{*} \left( {{\upzeta }_{1} = {\upzeta }_{2} = 0} \right) = F_{1} \left( \varDelta \right) \triangleq \frac{{\varDelta \left( {4 + \varDelta } \right)\left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} }} - F$. For supplier 2, the benefit from certification can be denoted as $\pi_{s2}^{*} \left( {{\upzeta }_{1} = {\upzeta }_{2} = 1} \right) - \pi_{s2}^{*} \left( {{\upzeta }_{1} = {\upzeta }_{2} = 0} \right) = F_{2} \left( \varDelta \right) \triangleq \frac{{\varDelta \left( {1 - \theta } \right)\left( {\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }} - F$. When $F > \hbox{max} \left\{ {\frac{{\varDelta \left( {4 + \varDelta } \right)\left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} }},\frac{{\varDelta \left( {1 - \theta } \right)\left( {\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }}} \right\} = \frac{{\varDelta \left( {1 - \theta } \right)\left( {\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }}$ in the 1-1 strategy, both suppliers will fall into the prisoner’s dilemma. In Cases (3)–(7), both suppliers will both certify the product when $F < \bar{F}_{13}$. Since $\bar{F}_{13} - \frac{{\varDelta \left( {1 - \theta } \right)\left( {\varDelta + \theta } \right)}}{{\left( {4 - \theta } \right)^{2} \theta }} = \frac{{\varDelta \left( {4\left( {1 - \theta } \right)\theta + \varDelta \left( { - 4 + \left( {4 - \theta } \right)\left( {2 - \theta } \right)\theta } \right)} \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$, if $\varDelta < \frac{{4\left( {1 - \theta } \right)\theta }}{{4 - \left( {4 - \theta } \right)\left( {2 - \theta } \right)\theta }}$, we get that both suppliers will fall into the prisoner’s dilemma if they reach 1-1 equilibrium in Cases (3)–(7). Since $\bar{F}_{21} - F_{2} \left( \varDelta \right) - F = \frac{{\left( {1 - \theta } \right)\theta }}{{4\left( { - 4 + \theta } \right)^{2} }} > 0$, we get that both suppliers will fall into the prisoner’s dilemma if they reach 1-1 equilibrium in Cases (8) and (10). In Cases (1) and (2), both suppliers will certify the product when $F < \bar{F}_{11}$. Since $\bar{F}_{11} - F_{2} \left( \varDelta \right) - F = \frac{{\varDelta \theta \left( {2 - \varDelta - 2\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} > 0$, we get that both suppliers will fall into the prisoner’s dilemma if they reach 1-1 equilibrium in Cases (1) and (2). In Case (9), both suppliers will certify the product when $F < \bar{F}_{33}$. Since $\bar{F}_{33} - F_{2} \left( \varDelta \right) - F = \frac{{\left( {1 - \theta } \right)\left( {\varDelta^{2} \left( { - 4 + \theta } \right) + 4\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \theta }}$, if $\varDelta < 2\sqrt {\frac{\theta }{4 - \theta }}$, we get that both suppliers will fall into the prisoner’s dilemma if they reach 1-1 equilibrium in Case (9).

Proof of Proposition 5

We then consider the condition that $1 - \theta < \varDelta \le { \hbox{min} }\left\{ {2 - 2\theta ,1} \right\}$; the retailer will only procure the products from supplier 1 in the certification–no-certification scenario. By equating $\pi_{s2yy}^{*}$ and $\pi_{s2yn}^{*}$, we get the threshold certification cost $\bar{F}_{21} = \frac{{\left( {2\varDelta + \theta } \right)^{2} \left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \theta }}$; by equating $\pi_{s2ny}^{*}$ and $\pi_{s2nn}^{*}$, we get $\bar{F}_{12} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta } \right)}}{{4\left( { - 4 + \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$; by equating $\pi_{s1yy}^{*}$ and $\pi_{s1ny}^{*}$, we get $\bar{F}_{13} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {4 - \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( { - 4 + \theta } \right)^{2} \left( {1 - \theta } \right)}}$; by equating $\pi_{s1yn}^{*}$ and $\pi_{s1nn}^{*}$, we get $\bar{F}_{22} = \frac{{\left( {1 + \Delta } \right)^{2} }}{16} - \frac{{\left( {1 - \theta } \right)}}{{\left( {4 - \theta } \right)^{2} }}$.

When $1 - \theta < \varDelta \le { \hbox{min} }\left\{ {2 - 2\theta ,1} \right\}$, we get that:

$\bar{F}_{13} - \bar{F}_{22} = \frac{{ - 2\varDelta \left( {1 - \theta } \right)\theta^{2} - \left( {1 - \theta } \right)\theta \left( {8 + \theta } \right) + \varDelta^{2} \left( { - 16 + \theta \left( {16 + \left( { - 5 + \theta } \right)\theta } \right)} \right)}}{{16\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} < 0$,

$\bar{F}_{21} - \bar{F}_{12} = \frac{{\varDelta^{2} \left( {4 - 3\theta } \right) + 2\varDelta \left( {1 - \theta } \right)\theta - \left( {1 - \theta } \right)^{2} \theta }}{{4\left( {4 - \theta } \right)^{2} \left( { - 1 + \theta } \right)}} < 0$ and $\frac{{\bar{F}_{12} }}{{\bar{F}_{13} }} = \frac{{\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta }}{{4\theta - \left( {4 + \varDelta } \right)\theta^{2} }} > 1$.

Since $\frac{{\bar{F}_{12} }}{{\bar{F}_{22} }} = \frac{{4\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta } \right)}}{{\left( {1 + \varDelta } \right)^{2} \left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta - 16\theta \left( {1 - \theta } \right)^{2} }}$, we let $F_{11} \left( {\theta ,\varDelta } \right) = 4\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + 2\left( {1 - \theta } \right)\theta } \right) - \left( {1 + \varDelta } \right)^{2} \left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta + 16\theta \left( {1 - \theta } \right)^{2}$, $\theta_{21} \approx 0.39$ and satisfies $F_{11} \left( {\theta_{21} ,\varDelta = 1 - \theta_{21} } \right) = 0$, $\theta_{22} \approx 0.5803$ and satisfies $F_{11} \left( {\theta_{22} ,\varDelta = 2 - 2\theta_{22} } \right) = 0$, $F_{11} \left( {\theta ,\varDelta = 1} \right) > 0$, when $\theta > \theta_{22}$, $\bar{F}_{12} < \bar{F}_{22}$, when $\theta < \theta_{21}$, $\bar{F}_{12} > \bar{F}_{22}$, when $\theta_{21} < \theta < \theta_{22}$, there exists $F_{11} \left( {\theta ,\varDelta_{21} } \right) = 0$ and satisfies $\varDelta_{21} = \frac{{\left( {1 - \theta } \right)\theta \left( {8 - \left( {4 - \theta } \right)\theta } \right) + 2\sqrt {\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta^{2} \left( {12 - \left( {4 - \theta } \right)\theta \left( {5 - 2\theta } \right)} \right)} }}{{16 - \left( {4 - \theta } \right)\theta \left( {8 - \left( {5 - \theta } \right)\theta } \right)}}$, if $\varDelta < \varDelta_{21}$, $\bar{F}_{12} < \bar{F}_{22}$, else, $\bar{F}_{12} > \bar{F}_{22}$.

Since $\bar{F}_{13} - \bar{F}_{21} = \frac{{ - \left( {1 - \theta } \right)^{2} \left( {2\varDelta + \theta } \right)^{2} - \varDelta \left( {2 - \theta } \right)\theta \left( { - 4 + \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \theta \left( {1 - \theta } \right)}}$, we let $F_{12} \left( {\theta ,\varDelta } \right) = - \left( {1 - \theta } \right)^{2} \left( {2\varDelta + \theta } \right)^{2} - \varDelta \left( {2 - \theta } \right)\theta \left( { - 4 + \left( {4 + \varDelta } \right)\theta } \right)$, and satisfies $F_{12} \left( {\frac{{5 - \sqrt {17} }}{2},\varDelta = 1 - \theta } \right) = 0$, $F_{12} \left( {\frac{{9 - \sqrt {17} }}{8},\varDelta = 2 - 2\theta } \right) = 0$, when $\theta > \frac{{9 - \sqrt {17} }}{8}$, $\bar{F}_{13} > \bar{F}_{21}$, when $\theta < \frac{{5 - \sqrt {17} }}{2}$, $\bar{F}_{13} < \bar{F}_{21}$, when $\frac{{5 - \sqrt {17} }}{2} < \theta < \frac{{9 - \sqrt {17} }}{8}$, there exists $\varDelta_{22} = \frac{{ - 2\left( {1 - \theta } \right)\theta - \sqrt {\left( {4 - \theta } \right)\left( {2 - \theta } \right)\left( {1 - \theta } \right)^{2} \theta^{3} } }}{{ - 4 + \left( {4 - \theta } \right)\left( {2 - \theta } \right)\theta }}$, if $\varDelta < \varDelta_{22}$, $\bar{F}_{13} > \bar{F}_{21}$, else, $\bar{F}_{13} < \bar{F}_{21}$.

Since $\bar{F}_{21} - \bar{F}_{22} = \frac{{4\left( {2\varDelta + \theta } \right)^{2} \left( {1 - \theta } \right) - \left( {4 - \theta } \right)^{2} \theta \left( {1 + \varDelta } \right)^{2} + 16\theta \left( {1 - \theta } \right)}}{{16\left( {4 - \theta } \right)^{2} \theta }}$, we let $F_{13} \left( {\theta ,\varDelta } \right) = 4\left( {2\varDelta + \theta } \right)^{2} \left( {1 - \theta } \right) - \left( {4 - \theta } \right)^{2} \theta \left( {1 + \varDelta } \right)^{2} + 16\theta \left( {1 - \theta } \right)$, $\theta_{23} \approx 0.288$ and satisfies $F_{13} \left( {\theta_{23} ,\varDelta = 1 - \theta_{23} } \right) = 0$, $\theta_{24} \approx 0.336$ and satisfies $F_{13} \left( {\theta_{24} ,\varDelta = 1} \right) = 0$, when $\theta > \theta_{24}$, $\bar{F}_{21} < \bar{F}_{22}$, when $\theta < \theta_{23}$, $\bar{F}_{21} > \bar{F}_{22}$, when $\theta_{23} < \theta < \theta_{24}$, there exists $F_{13} \left( {\theta ,\varDelta_{23} } \right) = 0$ and satisfies $\varDelta_{23} = - \frac{{8\theta + \theta^{3} + 2\theta \sqrt {\left( {1 - \theta } \right)\left( {32 + \theta \left( {20 - \left( {8 - \theta } \right)\theta } \right)} \right)} }}{{ - 16 + \theta \left( {32 - \left( {8 - \theta } \right)\theta } \right)}}$, if $\varDelta < \varDelta_{23}$, $\bar{F}_{21} < \bar{F}_{22}$, else, $\bar{F}_{21} > \bar{F}_{22}$.

In conclusion, we obtain the results in Table 3.

We then consider the condition that $2 - 2\theta < \Delta \le 1 \cap \theta > \frac{1}{2}$, thus, when in the certification–no-certification scenario, the retailer will only procure the products from supplier 1; when in the no-certification–certification scenario, the retailer will only procure the products from supplier 2.

By equating $\pi_{s2yy}^{*}$ and $\pi_{s2yn}^{*}$, we get the threshold certification cost $\bar{F}_{21} = \frac{{\left( {2\varDelta + \theta } \right)^{2} \left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \theta }}$; by equating $\pi_{s2ny}^{*}$ and $\pi_{s2nn}^{*}$, we get $\bar{F}_{34} = \frac{{\left( {\theta + \Delta } \right)^{2} }}{16\theta } - \frac{{\theta \left( {1 - \theta } \right)}}{{4\left( { - 4 + \theta } \right)^{2} }}$; by equating $\pi_{s1yy}^{*}$ and $\pi_{s1ny}^{*}$, we get $\bar{F}_{33} = \frac{{\left( {1 - \theta } \right)\left( {\varDelta + 2} \right)^{2} }}{{4\left( {4 - \theta } \right)^{2} }}$; by equating $\pi_{s1yn}^{*}$ and $\pi_{s1nn}^{*}$, we get $\bar{F}_{22} = \frac{{\left( {1 + \varDelta } \right)^{2} }}{16} - \frac{1 - \theta }{{\left( {4 - \theta } \right)^{2} }}$. Then, we get $\bar{F}_{21} - \bar{F}_{34} = \frac{ - \theta }{{16\left( {4 - \theta } \right)^{2} \theta }}\left( {8\theta + \theta^{3} + \varDelta^{2} \left( {8 + \theta } \right) + 2\varDelta \left( {8 + \theta^{2} } \right)} \right) < 0$, $\bar{F}_{33} - \bar{F}_{22} = \frac{{16 - 4\varDelta \left( {4 + 3\varDelta } \right) - 24\theta + 4\varDelta^{2} \theta - \left( {1 + \varDelta } \right)^{2} \theta^{2} }}{{16\left( {4 - \theta } \right)^{2} }} < 0$. Since $\bar{F}_{21} - \bar{F}_{33} = \frac{{\left( {\varDelta^{2} - \theta } \right)\left( {4 - \theta } \right)\left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \theta }}$ , thus if $\varDelta < \sqrt \theta$, $\bar{F}_{21} < \bar{F}_{33}$; if $\varDelta > \sqrt \theta$, $\bar{F}_{21} > \bar{F}_{33}$. $\bar{F}_{21} - \bar{F}_{22} = \frac{{ - \left( {1 + \varDelta } \right)^{2} \left( {4 - \theta } \right)^{2} \theta + 16\left( {1 - \theta } \right)\theta + 4\left( {1 - \theta } \right)\left( {2\varDelta + \theta } \right)^{2} }}{{16\theta \left( {4 - \theta } \right)^{2} }} < 0$, $\bar{F}_{34} - \bar{F}_{22} = \frac{{\left( {1 - \theta } \right)\left( {\varDelta^{2} \left( {4 - \theta } \right) + \theta^{2} } \right)}}{{16\theta \left( {4 - \theta } \right)}} > 0$. Thus, we get the results in Table 4.

Proof of Proposition 6

When both suppliers certify the product, we get that $\pi_{s1}^{*} - \pi_{s2}^{*} = \frac{{\left( {1 - \theta } \right)\left( { - \varDelta^{2} + \theta } \right)}}{{4\left( {4 - \theta } \right)\theta }}$, if $\varDelta < \sqrt \theta$, $\pi_{s1}^{*} > \pi_{s2}^{*}$; if $\varDelta > \sqrt \theta$, $\pi_{s1}^{*} < \pi_{s2}^{*}$. When neither supplier certifies the product, we get that $\pi_{s1}^{*} - \pi_{s2}^{*} = \frac{1 - \theta }{{4\left( {4 - \theta } \right)}} > 0$. To realize $\pi_{s1}^{*} < \pi_{s2}^{*}$, the condition $\theta < \frac{3 - \sqrt 5 }{2}$ and $\sqrt \theta < \Delta < 1 - \theta$ must be satisfied. We find $\varDelta_{11} < \varDelta_{12} < \sqrt \theta$, but there exists $\theta_{31}$ satisfying $\varDelta_{13} \left( {\theta_{31} } \right) = \sqrt {\theta_{31} }$, $\theta_{31} \approx 0.263$,$\varDelta_{13} > \sqrt \theta$ if $\theta_{31} < \theta < \frac{{7 - \sqrt {41} }}{2}$. Thus, if the two suppliers reach 1-1 equilibrium, supplier 2 may obtain more profits than supplier 1 if $\theta < \frac{{7 - \sqrt {41} }}{2}$ in Cases (3) and (4), and if $\frac{{7 - \sqrt {41} }}{2} < \theta < \frac{{5 - \sqrt {17} }}{2}$ in Case (3). When $\bar{F}_{13} < F < \bar{F}_{12}$. in Cases (2)–(4), the two suppliers achieve 0–1 equilibrium, we get $\pi_{s1}^{*} - \pi_{s2}^{*} = F - \frac{{\varDelta^{2} + 2\varDelta \theta - \theta \left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)\theta }}$. Since $f_{6} \left( \Delta \right) = \frac{{\varDelta^{2} + 2\varDelta \theta - \theta \left( {1 - \theta } \right)}}{{4\left( {4 - \theta } \right)\theta }} - \bar{F}_{13} = \frac{1}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}\left( {\left( { - 4 + \theta } \right)\left( {1 - \theta } \right)^{2} \theta - 2\varDelta \left( {1 - \theta } \right)\theta^{2} + \varDelta^{2} \left( {4 - \theta \left( {5 - \left( {3 - \theta } \right)\theta } \right)} \right)} \right)$, if $f_{6} \left( \Delta \right) > 0$, we get $\pi_{s1}^{*} < \pi_{s2}^{*}$ in a 0–1 equilibrium. $f_{6} \left( {\theta_{32} ,\Delta = 1 - \theta_{32} } \right) = 0$, $\theta_{32} \approx 0.484$, if $\theta > \theta_{32}$, $f_{6} \left( {\Delta = 1 - \theta } \right) < 0$, thus, $f_{6} \left( \Delta \right) < 0$. $f_{6} \left( {\Delta = \varDelta_{11} } \right) < 0$, $f_{6} \left( {\theta_{33} ,\Delta = \varDelta_{12} } \right) = 0$, $\theta_{33} \approx 0.403$, if $\theta > \theta_{33}$, $f_{6} \left( {\Delta = \varDelta_{12} } \right) > 0$, else, $f_{6} \left( {\Delta = \varDelta_{12} } \right) < 0$. $f_{6} \left( {\theta_{34} ,\Delta = \varDelta_{13} } \right) = 0$, $\theta_{34} \approx 0.243$, if $\theta > \theta_{34}$, $f_{6} \left( {\Delta = \varDelta_{13} } \right) > 0$, else, $f_{6} \left( {\Delta = \varDelta_{13} } \right) < 0$. Thus, we get there exists $\theta$ satisfying $\pi_{s1}^{*} < \pi_{s2}^{*}$ if $\theta < \frac{{7 - \sqrt {41} }}{2}$ in Cases (3) and (4), $\frac{{7 - \sqrt {41} }}{2} < \theta < \frac{{5 - \sqrt {17} }}{2}$ in Cases (2) and (3) and if $\frac{{5 - \sqrt {17} }}{2} < \theta < \theta_{32}$ in Case (2). When $\bar{F}_{11} < F < \bar{F}_{13}$ in Cases (1)–(3), the two suppliers achieve 1-0 equilibrium, we get $\pi_{s1}^{*} - \pi_{s2}^{*} = \frac{{\left( {1 + \varDelta } \right)^{2} - \theta }}{{4\left( {4 - \theta } \right)}} - F$. Since $f_{7} \left( \Delta \right) = \bar{F}_{13} - \frac{{\left( {1 + \varDelta } \right)^{2} - \theta }}{{4\left( {4 - \theta } \right)}} = \frac{1}{{4\left( {4 - \theta } \right)^{2} \left( { - 1 + \theta } \right)}}\left( {\varDelta^{2} \left( {4 - 3\theta } \right) + \left( {4 - \theta } \right)\left( {1 - \theta } \right)^{2} + 2\varDelta \left( {1 - \theta } \right)\theta } \right) < 0$, thus, $\pi_{s1}^{*} > \pi_{s2}^{*}$.

Appendix 4: The equilibrium certification strategy

Case (1) $\bar{F}_{11} < \bar{F}_{12} < \bar{F}_{13} < \bar{F}_{14}$

If $F \le \bar{F}_{11}$, supplier 1(2) always chooses certification regardless of the choice of supplier 2(1); thus, they will reach the certification–certification Nash equilibrium. To simplify the exposition, we denote the equilibrium as 1(dominant)-1(dominant); if $\bar{F}_{11} < F \le \bar{F}_{12}$, supplier 1 always chooses certification regardless of the choice of supplier 2, and supplier 2 will choose no certification if supplier 1 chooses certification; thus, they will reach the certification–no-certification Nash equilibrium, we denote the equilibrium as 1(dominant)-0. If $\bar{F}_{12} < F \le \bar{F}_{13}$, they will reach 1(dominant)-0 (dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{14}$, they will reach 1-0(dominant) Nash equilibrium; if $F > \bar{F}_{14}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (2) $\bar{F}_{11} < \bar{F}_{13} < \bar{F}_{12} < \bar{F}_{14}$

If $F \le \bar{F}_{11}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{11} < F \le \bar{F}_{13}$, they will reach 1(dominant)-0 Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{12}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{12} < F \le \bar{F}_{14}$, they will reach 1-0 (dominant) Nash equilibrium; if $F > \bar{F}_{14}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (3) $\bar{F}_{13} < \bar{F}_{11} < \bar{F}_{14} < \bar{F}_{12}$

If $F \le \bar{F}_{13}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{11}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{11} < F \le \bar{F}_{14}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{14} < F \le \bar{F}_{12}$, they will reach 0 (dominant)-1 Nash equilibrium; if $F > \bar{F}_{12}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (4) $\bar{F}_{13} < \bar{F}_{14} < \bar{F}_{11} < \bar{F}_{12}$

If $F \le \bar{F}_{13}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{14}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{14} < F \le \bar{F}_{11}$, they will reach 0(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{11} < F \le \bar{F}_{12}$, they will reach 0(dominant)-1 Nash equilibrium; if $F > \bar{F}_{12}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (5) $\bar{F}_{13} < \bar{F}_{22} < \bar{F}_{21} < \bar{F}_{12}$

If $F \le \bar{F}_{13}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{22}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{22} < F \le \bar{F}_{21}$, they will reach 0(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{12}$, they will reach 0 (dominant)-1 Nash equilibrium; if $F > \bar{F}_{12}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (6) $\bar{F}_{13} < \bar{F}_{21} < \bar{F}_{22} < \bar{F}_{12}$

If $F \le \bar{F}_{13}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{21}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{22}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{22} < F \le \bar{F}_{12}$, they will reach 0 (dominant)-1 Nash equilibrium; if $F > \bar{F}_{12}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (7) $\bar{F}_{13} < \bar{F}_{21} < \bar{F}_{12} < \bar{F}_{22}$

If $F \le \bar{F}_{13}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{21}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{12}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{12} < F \le \bar{F}_{22}$, they will reach 1-0 (dominant) Nash equilibrium; if $F > \bar{F}_{22}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (8) $\bar{F}_{21} < \bar{F}_{13} < \bar{F}_{12} < \bar{F}_{22}$

If $F \le \bar{F}_{21}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{13}$, they will reach 1(dominant)-0 Nash equilibrium; if $\bar{F}_{13} < F \le \bar{F}_{12}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{12} < F \le \bar{F}_{22}$, they will reach 1-0 (dominant) Nash equilibrium; if $F > \bar{F}_{22}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (9) $\bar{F}_{33} < \bar{F}_{21} < \bar{F}_{22} < \bar{F}_{34}$

If $F \le \bar{F}_{33}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{33} < F \le \bar{F}_{21}$, they will reach 0–1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{22}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{22} < F \le \bar{F}_{34}$, they will reach 0 (dominant)-1 Nash equilibrium; if $F > \bar{F}_{34}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Case (10) $\bar{F}_{21} < \bar{F}_{33} < \bar{F}_{22} < \bar{F}_{34}$

If $F \le \bar{F}_{21}$, they will reach 1(dominant)-1(dominant) Nash equilibrium; if $\bar{F}_{21} < F \le \bar{F}_{33}$, they will reach 1(dominant)-0 Nash equilibrium; if $\bar{F}_{33} < F \le \bar{F}_{22}$, they will reach 1–0 or 0–1 Nash equilibrium; if $\bar{F}_{22} < F \le \bar{F}_{34}$, they will reach 0 (dominant)-1 Nash equilibrium; if $F > \bar{F}_{34}$, they will reach 0(dominant)-0(dominant) Nash equilibrium.

Each of the cases above can be divided into 5 parts with the increase of $F$; we denote five parts in each case as the first part, second part–fifth part, respectively, showing that the two suppliers will reach 1-1 equilibrium in the first part, and reach 0-0 equilibrium in the fifth part.

Appendix 5: The certification and pricing decisions, considering production cost

We incorporate the production cost into the model, and analyze the certification and pricing decisions in a simultaneous game. We mainly focus on the condition when the retailer will always procure two products regardless of their certification strategy.

When $\frac{{p_{2} - \upzeta_{2} \Delta }}{{p_{1} - \upzeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\upzeta_{1} - \upzeta_{2} } \right)\Delta$, we get the optimal pricing decisions,

$$ \begin{aligned} m_{1}^{*} & = \frac{1}{2}\left( {1 - c + \varDelta \zeta_{1} } \right), \\ m_{2}^{*} & = \frac{1}{2}\left( {\theta + \varDelta \zeta_{2} - c} \right), \\ w_{1}^{*} & = \frac{{2 + 7c - \left( {2 + c} \right)\theta + \varDelta \left( {2 - \theta } \right)\zeta_{1} - \varDelta \zeta_{2} }}{{2\left( {4 - \theta } \right)}}, \\ w_{2}^{*} & = \frac{{6c + \left( {1 - \theta } \right)\theta - \varDelta \theta \zeta_{1} + \varDelta \left( {2 - \theta } \right)\zeta_{2} }}{{2\left( {4 - \theta } \right)}}, \\ p_{1}^{*} & = \frac{{6 + 3c - 3\theta + 2\varDelta \left( {3 - \theta } \right)\zeta_{1} - \varDelta \zeta_{2} }}{{2\left( {4 - \theta } \right)}}, \\ p_{2}^{*} & = \frac{{c\left( {2 + \theta } \right) + \theta \left( {5 - 2\theta } \right) - \varDelta \theta \zeta_{1} + 2\varDelta \left( {3 - \theta } \right)\zeta_{2} }}{{2\left( {4 - \theta } \right)}}, \\ \pi_{s1}^{*} & = \frac{{\left( {\left( {2 - c} \right)\left( {1 - \theta } \right) + \varDelta \left( {2 - \theta } \right)\zeta_{1} - \varDelta \zeta_{2} } \right)^{2} }}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}} - \upzeta_{1} F, \\ \pi_{s2}^{*} & = \frac{{\left( {\left( {2c - \theta } \right)\left( {1 - \theta } \right) + \varDelta \theta \zeta_{1} - \varDelta \left( {2 - \theta } \right)\zeta_{2} } \right)^{2} }}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }} - \upzeta_{2} F, \\ \pi_{r}^{*} & = \frac{1}{{4\left( {4 - \theta } \right)\left( {1 - \theta } \right)\theta }}\left( {\left( {1 - \theta } \right)\left( {\left( {c^{2} + \theta } \right)\left( {2 + \theta } \right) - 6c\theta } \right) + \varDelta^{2} \left( {2 - \theta } \right)\theta \zeta_{1}^{2} + 2\varDelta \left( {1 - \theta } \right)\left( {\theta - 2c} \right)\zeta_{2} } \right. \\ & \quad \quad \left. { + \varDelta^{2} \left( {2 - \theta } \right)\zeta_{2}^{2} + 2\varDelta \theta \zeta_{1} \left( {\left( {2 - c} \right)\left( {1 - \theta } \right) - \varDelta \zeta_{2} } \right)} \right). \\ \end{aligned} $$

After substituting $p_{1}^{ *}$ and $p_{2}^{ *}$ into the constraint $\frac{{p_{2} - \upzeta_{2} \Delta }}{{p_{1} - \upzeta_{1} \Delta }} \le \theta \le 1 - p_{1} + p_{2} + \left( {\upzeta_{1} - \upzeta_{2} } \right)\Delta$, we get the following conditions under which the retailer will simultaneously procure two products.

In scenario 1, if $c \le \frac{2\varDelta + \theta }{2}$, the retailer will procure two products simultaneously, otherwise, he will only procure product 1; in scenario II, if $c \le \frac{{\theta \left( {1 - \varDelta - \theta } \right)}}{{2\left( {1 - \theta } \right)}}$, the retailer will procure two products simultaneously, otherwise, he will only procure product 1; in scenario III, if $\varDelta \ge 1 - \theta$ and $c \le \frac{{2\left( {2 - \varDelta - 2\theta } \right)}}{{2\left( {1 - \theta } \right)}}$, the retailer will procure two products simultaneously, otherwise, he will only procure product 2; if $\varDelta \le 1 - \theta$ and $c \le \frac{{2\varDelta + \theta - \theta \left( {\varDelta + \theta } \right)}}{{2\left( {1 - \theta } \right)}}$, the retailer will procure two products simultaneously, otherwise, he will only procure product 1; in scenario IV, if $c \le \frac{\theta }{2}$, the retailer will procure two products simultaneously, otherwise, he will only procure product 1. We mainly focus on the condition of $\varDelta \le 1 - \theta - \frac{{2\left( {1 - \theta } \right)c}}{\theta }$, and we get $c \le \frac{\theta }{2}$ according to $1 - \theta - \frac{{2\left( {1 - \theta } \right)c}}{\theta } \ge 0$.

For supplier 2, given that supplier 1 adopts certification, supplier 2 adopts certification if $F \le \bar{F}_{41} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$, otherwise no certification; given that supplier 1 adopts no certification, supplier 2 adopts certification if $F \le \bar{F}_{42} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$, otherwise no certification. For supplier 1, given that supplier 2 adopts certification, supplier 1 adopts certification if $F \le \bar{F}_{43} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {4 - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( { - 4 + \theta } \right)^{2} \left( {1 - \theta } \right)}}$, otherwise no certification; given that supplier 2 adopts no certification, supplier 1 adopts certification if

$F \le \bar{F}_{44} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {2\left( {2 + \varDelta } \right) - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)}}$, otherwise no certification.

$\frac{{\bar{F}_{41} }}{{\bar{F}_{42} }} = \frac{{\varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)}}{{\varDelta \left( {2 - \theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)}} < 1,\quad \frac{{\bar{F}_{43} }}{{\bar{F}_{44} }} = \frac{{4 - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta }}{{2\left( {2 + \varDelta } \right) - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta }} < 1.$ Since $\frac{{\bar{F}_{41} }}{{\bar{F}_{44} }} = \frac{{\left( {\varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{\theta \left( {2\left( {2 + \varDelta } \right) - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}$, we let $F_{41} \left( {\theta ,\varDelta } \right) = \varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right) - \theta \left( {2\left( {2 + \varDelta } \right) - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)$. Since $\frac{{\bar{F}_{42} }}{{\bar{F}_{43} }} = \frac{{\varDelta \left( {2 - \theta } \right)\left( {\varDelta \left( {2 - \theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{4\left( {4 - \theta } \right)^{2} \left( {1 - \theta } \right)\theta }}$, we let $F_{42} \left( {\theta ,\varDelta } \right) = \varDelta \left( {2 - \theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right) - \theta \left( {4 - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)$. Since $\frac{{\bar{F}_{41} }}{{\bar{F}_{43} }} = \frac{{\left( {\varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{\theta \left( {4 - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}$ and $\frac{{\bar{F}_{42} }}{{\bar{F}_{44} }} = \frac{{\left( {\varDelta \left( {2 - \theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)} \right)}}{{\theta \left( {2\left( {2 + \varDelta } \right) - 2c\left( {1 - \theta } \right) - \left( {4 + \varDelta } \right)\theta } \right)}}$, we let $F_{43} \left( {\theta ,\varDelta } \right) = \varDelta \left( {2 - 3\theta } \right) + \left( {2\theta - 4c} \right)\left( {1 - \theta } \right)$, we let $F_{41} \left( {\theta ,1 - \theta - \frac{{2\left( {1 - \theta } \right)c_{1} }}{\theta }} \right) = 0$ and $c_{1} = \frac{{\theta \left( {\theta^{2} - 7\theta + 2} \right)}}{4 - 6\theta }$; $F_{42} \left( {\theta ,1 - \theta - \frac{{2\left( {1 - \theta } \right)c_{2} }}{\theta }} \right) = 0$ and $c_{2} = \frac{{\theta \left( {2 - \theta } \right)\left( {1 - \theta } \right)}}{{2\left( {2 + \theta } \right)}}$; $F_{43} \left( {\theta ,1 - \theta - \frac{{2\left( {1 - \theta } \right)c_{3} }}{\theta }} \right) = 0$ and $c_{3} = \frac{{\theta \left( {\theta^{2} - 5\theta + 2} \right)}}{{2\left( {2 - \theta } \right)}}$; $F_{41} \left( {\theta ,\varDelta_{41} } \right) = 0$ and $\varDelta_{41} = \frac{{2\left( {c\left( {2 - \theta } \right) + \theta } \right)\left( {1 - \theta } \right)}}{{2 - \left( {1 - \theta } \right)\theta }}$; $F_{42} \left( {\theta ,\varDelta_{42} } \right) = 0$ and $\varDelta_{42} = 2c + \frac{2\theta }{2 - \theta }$; $F_{43} \left( {\theta ,\varDelta_{43} } \right) = 0$ and $\varDelta_{43} = \frac{{2\left( {c\left( {2 - \theta } \right) + \theta } \right)\left( {1 - \theta } \right)}}{{2 - \left( {5 - \theta } \right)\theta }}$, $\varDelta_{41} < \varDelta_{42} < \varDelta_{43}$, then, we get Table 5.

Table 5 The organic certification strategy when $\varDelta \le 1 - \theta - \frac{{2\left( {1 - \theta } \right)c}}{\theta }$

(11) $\bar{F}_{41} < \bar{F}_{42} < \bar{F}_{43} < \bar{F}_{44}$, the certification strategy here is similar to that in Case (1); (12) $\bar{F}_{41} < \bar{F}_{43} < \bar{F}_{42} < \bar{F}_{44}$, the certification strategy here is similar to that in Case (2); (13) $\bar{F}_{43} < \bar{F}_{41} < \bar{F}_{44} < \bar{F}_{12}$, the certification strategy here is similar to that in Case (3); (14) $\bar{F}_{13} < \bar{F}_{14} < \bar{F}_{11} < \bar{F}_{12}$, the certification strategy here is similar to that in Case (4).